Problem: What is the slope of the line tangent to $f(x) = -2x^{2}+4x+3$ at $x = 1$ ?
Explanation: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-2(x+h)^{2}+4(x+h)+3) - (-2x^{2}+4x+3)}{h}$ $ = \lim_{h \to 0} \frac{(-2(x^{2}+2x h+h^{2})+4(x+h)+3) - (-2x^{2}+4x+3)}{h}$ $ = \lim_{h \to 0} \frac{-2x^{2}-4(x h)-2h^{2}+4x+4h+3+2x^{2}-4x-3}{h}$ $ = \lim_{h \to 0} \frac{-4(x h)-2h^{2}+4h}{h}$ $ = \lim_{h \to 0} -4x-2h+4$ $ = -4x+4$ $ = (-4)(1)+4$ $ = 0$